3) With a headwind, a plane traveled 840 miles northward in 2 hours. With the same wind as a tail wind, the return trip took 1 hour and 45 mineutes. Find the plane's airspeed and the wind speed.

for 3), I attempted this problem by simply making a graph... I got confused...

4) With a tailwind, a small plane traveled 420 miles in one hour and 30 min. The return trip took 30 min longer. Find the wind speed and airspeed of the plane...

for 4) I plugged in random numbers for both... I think I made the wrong equation... PLS HALP

tommarvoloriddle Jun 25, 2019

#1**0 **

Rate x time = distance

Rate = Speed + wind or speed - wind

speed-wind * 2 = 840

Speed + wind * 1.45 = 840 Two equations with two unknowns

From first equation W+420 =S Sub in to the second equation

(W+420+W)*1.45 = 840

W= 30 mph

Sub this result in to one of the equations to solve for S

S-30 * 2 = 840 S = 450 mph

ElectricPavlov Jun 26, 2019

#2**0 **

**EP: How did you get: **

**(W+420+W)*1.45 = 840**

**W= 30 mph ???**

**1 hour and 45 minutes =1 hour + 45/60 =1.75 -An hour and three-quarters. Why did you multiply by 1.45 ???.**

**Your W works out =2310 /29 =~79.65 ???**

Guest Jun 26, 2019

#7**0 **

That was just a typo 1.45 should be 1.75 or 1:45 answer does not change....when I calculated it I used 1.75 but I typed 1.45

ElectricPavlov
Jun 26, 2019

#3**+3 **

Perhaps the following will make it clearer:

EP made the trivial mistake of setting 1 hour 45 mins (i.e. one and three quarter hours) as 1.45 hours rather than 1.75 hours

Now see how you get on with your question 4) .

Alan Jun 26, 2019

#4**+1 **

3) With a headwind, a plane traveled 840 miles northward in 2 hours. With the same wind as a tail wind, the return trip took 1 hour and 45 mineutes. Find the plane's airspeed and the wind speed.

Note....1hr 45 min = 1.75 hrs

Let the speed of the plane in still air be S

Let the wind speed be W

Against the wind the plane's rate is (R - W) ....the wind slows the plane's normal speed

With the wind the plane's rate is (R + W).....the wind increases the plane's normal speed

So..... Rate * Time = Distance and we have that

(R -W) * 2 = 840 ⇒ 2R - 2W = 840 ⇒ R - W = 420 ⇒ R = 420 + W (1)

(R + W) *1.75 = 840 ⇒ 1.75 R + 1.75W = 840 (2)

Sub (1) into (2) for R and we have that

1.75(420 + W) + 1.75W = 840

735 + 1.75W + 1.75W = 840

735 + 3.5W = 840 subtract 735 from both sides

3.5W = 105 divide both sides by 3.5

W = 30 mph = wind speed

And using (1) the plane's speed is 420 + 30 = 450 mph

CPhill Jun 26, 2019

#5**+1 **

4) With a tailwind, a small plane traveled 420 miles in one hour and 30 min. The return trip took 30 min longer. Find the wind speed and airspeed of the plane...

This one is similar to (3)

1hr 30 min = 1.5 hrs.....so

(R + W) * 1.5 = 420 (1)

(R-W ) *2 = 420 ⇒ 2R - 2W = 420 ⇒ R - W = 210 ⇒ R = 210 + W (2)

Sub (2) into (1) for R and we have that

(210 + W + W) * 1.5 = 420

(210 + 2W) * 1.5 = 420

315 + 3W = 420 subtract 315 from each side

3W = 105 divide both sides by 3

W = 35 mph

And using (1) , R = 210 + 35 = 245 mph

CPhill Jun 26, 2019